Chapter 12.8, Problem 15PPS

(a)

To determine

To find: the scale factor between the large and small cheese wheels.

Answer to Problem 15PPS

The scale factor between the large and small cheese wheels is $\frac{4}{1}$ .

Explanation of Solution

Given:

The diameter of the cylindrical cheese is 48 in

The height of the cheese is 17 in.

The diameter of another cheese is 1 ft.

Calculation:

The objective is to find the scale factor between the large and small cheese wheels.

First convert feet into inches.

  $1\text{ ft}=12\text{ in}\text{.}$

Two solid figures are similar if they have the same shape and their corresponding linear measurements are proportional.

Since both the figures are cylinder and their corresponding linear measurements are proportional, so the cylinders are similar.

The ratio of the diameter of large to smaller cheese is $\frac{48}{12}$ that is $\frac{4}{1}$ .

Conclusion:

Therefore, the scale factor between the large and small cheese wheels is $\frac{4}{1}$ .

(b)

To determine

To find:the height of the smaller cheese wheel.

Answer to Problem 15PPS

The height of the smaller cheese wheel is 4.25 in..

Explanation of Solution

Calculation:

The objective is to find the height of the smaller cheese wheel.

Let h be the height of the smaller cheese wheel.

Since the cheese wheels are similar, so their corresponding linear measurements are proportional.

Thus,

  $\begin{array}{l}\frac{17}{h}=\frac{48}{12}\\ =\frac{4}{1}\\ \frac{17}{h}=\frac{4}{1}\\ 17\cdot 1=4h\\ 17=4h\\ 4h=17\\ h=\frac{17}{4}\\ =4.25\end{array}$

Conclusion:

Therefore, the height of the smaller cheese wheel is 4.25 in..

(c)

To determine

To find: volume of the each wheel of cheese.

Answer to Problem 15PPS

The volume of the larger wheel is approximately $2678394.88{\text{ in}}^3$ and the smaller wheel is $480.42i{n}^3$ .

Explanation of Solution

Calculation:

The objective is to find the volume of each wheel cheese.

The volume V of the cylinder is given by the formula

  $V=\pi r^{2}h$

Let $V_1$ be the volume of larger cylinder.

Thus,

  $\begin{array}{l}{V}_1=\pi {\left(24\right)}^{2}17\\ =\pi \cdot 50176\cdot 17\\ =852992\pi \\ \approx 852992\left(3.14\right)[Since\pi \approx \text{3}\text{.14}]\end{array}$

Therefore, the volume of the larger wheel is approximately $2678394.88{\text{ in}}^3$ .

Let $V_2$ be the volume of larger cylinder.

Thus,

  $\begin{array}{l}{V}_2=\pi {\left(6\right)}^2\left(4.25\right)\\ =\pi \cdot 36\left(4.25\right)\\ =153\pi \\ \approx 153\left(3.14\right)[Since\pi \approx \text{3}\text{.14}]\\ =480.42\end{array}$

Conclusion:

Therefore,the volume of the larger wheel is approximately $2678394.88{\text{ in}}^3$ and the smaller wheel is $480.42i{n}^3$ .

(d)

To determine

To find: the difference in the volume of larger wheel to smaller wheel.

Answer to Problem 15PPS

The difference is $2699914.46i{n}^3$ .

Explanation of Solution

Calculation:

The objective is to find the difference in the volume of larger wheel to smaller wheel.

The volume of larger wheel is $2678394.88i{n}^3$ and the volume smaller wheel is $480.42i{n}^3$ .

The difference is $\left(2678394.88-480.42\right)i{n}^3$ that is $2699914.46i{n}^3$

Conclusion:

Therefore,the difference is $2699914.46i{n}^3$ .